ch32-p035 - 35. (a) Since m = 0, Lorb,z = m h/2 = 0. (b)...

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(d) Regardless of the value of m A , we find for the spin part UB B sz B =− × × −− µµ , .. . 9 2 71 0 3 5 3 21 0 24 25 JT mT J c h bg (e) Now m A = –3, so () 27 34 34 orb, 36 . 6 31 0J s 3.16 10 J s 3.2 10 J s 22 z mh L −× = = = ×⋅ ππ A (f) and 24 23 23 orb, 3 9.27 10 J T 2.78 10 J T 2.8 10 J T . zB m =− − × = × × A (g) The potential energy associated with the electron’s orbital magnetic moment is now ( ) 23 3 25 orb, ext 2.78 10 J T 35 10 T 9.7 10 J. z µ −−− × × × (h) On the other hand, the potential energy associated with the electron spin, being
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